def longest_palindromic_subsequence(s):
    n = len(s)
    # dp[i][j] 表示字符串 s 的子串 s[i:j+1] 的最长回文子序列的长度
    dp = [[0] * n for _ in range(n)]
    
    # 对角线上的元素初始化为 1，因为单个字符的回文子序列长度为 1
    for i in range(n):
        dp[i][i] = 1
    
    # 填充 dp 表
    for cl in range(2, n+1):  # cl 是子串的长度
        for i in range(n - cl + 1):
            j = i + cl - 1
            if s[i] == s[j] and cl == 2:
                dp[i][j] = 2
            elif s[i] == s[j]:
                dp[i][j] = dp[i + 1][j - 1] + 2
            else:
                dp[i][j] = max(dp[i][j - 1], dp[i + 1][j])
    
    return dp[0][n - 1]

def make_palindrome(s):
    lps_length = longest_palindromic_subsequence(s)
    # 计算需要添加的字符数量
    add_chars_count = len(s) - lps_length
    # 构造回文串
    palindrome = s + s[:add_chars_count][::-1]
    return palindrome

# 测试
input_str = "abca"
print(make_palindrome(input_str))  # 输出: "abcaxba"
